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Consider a ladder slanted like \ of length ##l## slipping against a smooth wall and on a smooth floor.

I come to the contradiction that there must be a deceleration in the x direction but there is no force opposing the velocity of the ladder.

Its free-body diagram contains a rightward normal force at the top end and a downward weight at the center of gravity.

Let the coordinates of the top and bottom ends be ##(0, y)## and ##(x, 0)## respectively. We have ##x^2+y^2=l^2##.

Let the distance moved by the top and bottom ends in time ##\Delta t## be ##\Delta y## and ##\Delta x## respectively. We have ##(x+\Delta x)^2+(y-\Delta y)^2=l^2##.

Ignoring higher powers of ##\Delta x## and ##\Delta y## and also using the first equation, we get ##\frac{\Delta y}{\Delta x}=\frac{x}{y}##. Dividing both the numerator and denominator by ##\Delta t##, we get ##\frac{v_y}{v_x}=\frac{x}{y}##. This means that ##v_x>0## shortly after the slipping begins, and when the ladder just hits the floor, ##y=0## and thus ##v_x=0##. In other words, there is a deceleration in the x direction. But then there is no leftward force in the free-body diagram.

What's wrong?

I come to the contradiction that there must be a deceleration in the x direction but there is no force opposing the velocity of the ladder.

Its free-body diagram contains a rightward normal force at the top end and a downward weight at the center of gravity.

Let the coordinates of the top and bottom ends be ##(0, y)## and ##(x, 0)## respectively. We have ##x^2+y^2=l^2##.

Let the distance moved by the top and bottom ends in time ##\Delta t## be ##\Delta y## and ##\Delta x## respectively. We have ##(x+\Delta x)^2+(y-\Delta y)^2=l^2##.

Ignoring higher powers of ##\Delta x## and ##\Delta y## and also using the first equation, we get ##\frac{\Delta y}{\Delta x}=\frac{x}{y}##. Dividing both the numerator and denominator by ##\Delta t##, we get ##\frac{v_y}{v_x}=\frac{x}{y}##. This means that ##v_x>0## shortly after the slipping begins, and when the ladder just hits the floor, ##y=0## and thus ##v_x=0##. In other words, there is a deceleration in the x direction. But then there is no leftward force in the free-body diagram.

What's wrong?

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